Case study (4 Marks)

Question 14 Marks

Read the following text and answer the following questions on the basis of the same:
A manufacturing company manufactures toys, the company observed the following costs at different production levels,

Number of toys manufactured	Cost of raw material (₹)	Cost of Produc-tion (₹)	Cost of freight (₹)	Property tax (₹)	Salaries (₹)
100	800	2000	1000	5000	20000
150	1200	3000	1500	5000	20000
200	1600	4000	2000	5000	20000
250	2000	5000	2500	5000	20000
300	2400	6000	3000	5000	20000

Q.1. Which of the following is the fixed cost
(A) Number of toys manufactured (B) Cost of raw material
(C) Cost of production supply (D) Salaries

Q.2. Total cost C(x) of toys for 'x' units of production is
(A) $C(x)=8 x^2+36 x+25000$
(B) $C(x)=8 x^2+30 x+20000$
(C) $C(x)=38 x+25000$
(D) $C(x)=28 x+25000$

Q.3. If the company observes the price p' per unit of item sold p = 5000 - 10x where x' is the number of units sold. Then the revenue function R(x) is given by.
(A) $R(x)=3000 x-10 x^2$
(B) $R(p)=5000 p-10 x^2$
(C) $R(x)=5000-10 x^2$
(D) $R(p)=5000-10 p^2$

Q.4. The marginal revenue (MR) of the company is given by
(A) 5000-20x
(B) 5000-20p
(C) - 20x
(D) - 20p

Answer

(1) (D) Salaries
Explanation: Salary always comes under fixed cost for a company.

(2) (C) C(x) = 38x + 25000
Explanation: From the given table, we get
Cost of raw material per toy = ₹৪
Cost of production supply per toy = ₹20
Cost of freight per toy = ₹10
Fixed cost = property fax + Salaries
= 5000 + 20000
= ₹ 25000
Hence, Cost function,
C(x) = (8 + 20 + 10)x + 25000
C(x) = 38x + 25000

(3) (A) $R(x)=5000 x-10 x^2$
Explanation: Revenue,
R(x) =p.x
= (5000 - 10x)x
$=5000 x-10 x^2$

(4) (A) 5000-20x
Explanation: Marginal revenue
$MR =\frac{d R(x)}{d x}$
$=\frac{d}{d x}\left(5000 x-10 x^2\right)$
= 5000 - 20x

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Question 24 Marks

Read the following text and answer the following questions on the basis of the same:
A farmer has a piece of land. He observed that he got 600 units of fruits per tree by planting upto 25 trees and when 26 trees were grown, he received 15210 units of fruits, for 27 trees he ended up with 15390 fruits, for 28 trees he got 15540 fruits and this sequence of production of fruits continues in the same pattern as more trees, in excess of 25, were grown

Q.1. If 'x' more trees, in excess of 25 are grown, then the number of fruits produced per tree is
(A) 600-15x
(B) 600 + 15x
(C) 600x-15
(D) 600x + 15

Q.2. The product of entire garden if x' more trees, in excess of 25, are planted
(A) (25 + x)(600 + 15x)
(B) (25 - x)(600 - 15x)
(C) (25 + x)(600 - 15x)
(D) (25 + x)(15x - 600)

Q.3. The marginal production of the garden when 'x' more trees, in excess of 25, are
(A) 225 + 30x
(B) 225-30x
(C) 225x + 30
(D) 225x-30

Q.4. The critical point of producing 'x' more units of tree is
(A) 7 (B) 8 (C) 7.5 (D) 8.5

Answer

(1) (A) 600-15x
Explanation: When no. of trees = 25, then no. of units of fruits per tree = 600
When no. of trees = 26, then no. of units of fruits per tree = $\frac{15210}{26}$ = 585 = (600-15)
When no. of trees 27, then no. of units of per tree = $\frac{15390}{27}=570=(600-15 \times 2)$
When no. of trees 28, then no. of units of fruits per tree $=\frac{15540}{28}=555=(600-15 \times 3)$
Thus, when x more trees, in excess of 25 are grown then the no. of fruits produced per tree is 600 - 15x

(2) (C) (25 + x)(600 - 15x)
Explanation: We know that, if 'x' more tree in excess of 25 are grown, then the number of fruits produced per tree is 600-15x.
So, the production of entire garden, if 'x' more trees, in excess of 25, are planted is given by p = (25 + x)(600 - 15x)

(3) (B) 225-30x
Explanation: Since, P = (25 + x)(600 - 15x)
$\therefore$ Marginal production,
$\frac{d P}{d x}=(600-15 x)+(25+x)(-15)$
= (600 - 375) - 30x
= 225 - 30x

(4) (C) 7.5
Explanation: For critical point Put $\frac{d P}{d x}=0$
i.e 225 - 30x = 0
$\Rightarrow \quad x=\frac{225}{30}$
= 7.5

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Question 34 Marks

Read the following text and answer the following questions on the basis of the same: "
A group of people decided to start a new project related to print media. They all have decided to publish a newspaper which has only college related news. They named their newspaper 'The Collegiate Investigator'. This newspaper has fixed production cost of ₹ 70 per edition and marginal printing and distribution cost of ₹ 0.40/per copy. The collegiate Investigator sells for ₹0.50/per copy.

Q.1. The associated cost function is:
(A) C(x) = 0.4x + 70
(B) C(x) = 0.4x - 70
(C) C(x) = - 0.4 + 70x
(D) C(x) = - 0.4 - 70x

Q.2. The associated revenue function is:
(A) R(x) = 0.4x
(B) R(x) = 0.5x
(C) R(x) = 0.7x
(D) R(x) = 0.5x + 70

Q.3. The associated profit function is:
(A) P(x) = - 0.1x - 70
(B) P(x) = - 0.1x + 50
(C) P(x) = 0.1x - 70
(D) P(x) = - 0.1x + 70

Q.4. What profit (or loss) results from the sale of 500 copies of The Collegiate Investigator?
(A) 20 loss "
(B) 20 profit
(C) 70 loss
(D) 70 profit

Answer

(1) (A) C(x) = 0.4x + 70
Explanation: Since, the marginal cost is ₹ 0.40 per copy and the fixed costs are ₹ 70. So, the cost function is
C(x) = 0.4x + 70

(2) (B) R(x) = 0.5x
Explanation: Since, the selling price of newspaper is ₹ 0.50 per copy. So, the revenue function is R(x) = 0.5x

(3) (C) P(x) = 0.1x - 70
Explanation: Since, profit is defined to be revenue minus cost, the profit function is P(x) = 0.5x - (0.4x + 70) = 0.1x - 70

(4) (A) 20 loss
Explanation: The profit from selling 500 copies is P(500)
i.e P(500) = 0.1(500) - 70 = - 20
This means that if they sell 500 newspaper, it will result in a loss of ₹ 20

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Question 44 Marks

Read the following text and answer the following questions on the basis of the same:
The market research department of a company recommends manufacture and market a new toy-car. The financial department provides the following cost function (in rupees) C(x) = 7,000 + $2 x ; 0 \leq x \leq 10,000$ where ₹7,000 is the estimate of fixed costs and 2 is the estimate of variable cost per toy-car. The estimate of revenue function (in rupees) is$R(x)=x(10-0.001 x) ; 0 \leq x \leq 10,000$.

Q.1. The marginal cost function is:
(A) 2x
(B) 2
(C) 7000+2x
(D) 7000

Q.2. The marginal revenue function is:
(A) 10-0.002x
(B) x(10-0.001x
(C) -0.002x
(D) 10

Q.3. The marginal revenue at x = 2000, 5000, 7000
(A) 6,-4,0
(B) 0,-4,6
(C) 6,0,-4
(D) none of these

Q.4. The profit function P(x) is given by
(A) $0.001 x^2+8 x+7000$
(B) $0.001 x^2-8 x-7000$
(C) $-0.001 x^2+8 x-7000$
(D) $-0.001 x^2-8 x-7000$

Answer

(1) (B) 2
Explanation: The marginal cost is
C(x) = 7000 + 2x
C'(x) = 2
Since the is a constant, it cost an additional ₹ 2 to produce one more toy-car at any production level.

(2) (A) x(10-0.001x)
Explanation: The marginal revenue is
R'(x) = 10 - 0.002x

(3) (C) 6,0,-4
Explanation: For production level of x = 2000, 5000 and 7000, we have R(x) = 10 - 0.002x
R'(2000) = 6
R'(5000) = 0
R'(7000) = - 4

(4) (C) $-0.001 x^2+8 x-7000$
Explanation: The profit function is
P(x) = R(x) - C(x)
$=-0.001 x^2+8 x-7000$

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Question 54 Marks

Read the following text and answer the following questions on the basis of the same:
Four friends Rohan, Riya, Kartik and Mayank are playing number game. One by one they are giving the riddles to others based on the numbers. Now, its Riya's chance to give a number riddle. Riya said: "The sum of three positive numbers is 26. The second number is thrice as large as the first and sum of the squares of the number is least".

Q.1. If x, y, z are first, second and third number respectively, then which of the following is true.
(A) x + y + z = 26 and y = 3x
(B) x + y + z = 26 and x = 3y
(C) x + y + z = 26 and z = 3x
(D) x + y + z = 26 and y = 3z

Q.2. If S is the sum of the squares of the three numbers, then which of the following relation is true.
(A) $S=26 x^2+208 x+676$
(B) $S =26 x ^2-208 x -676$
(C) $5=26 x^2-208 x+675$
(D) $5=-26 x^2-208 x+676$

Q.3. The second derivative of S with respect to x is:
(A) 52
(B) 25
(C) 52x - 208
(D) 52x + 208

Q.4. At what values of x, the sum of the square of the numbers is least?
(A) 5
(B) 2
(C) 8
(D) 4

Answer

(1) (A) x + y + z = 26 and y = 3x
Explanation: If x, y, z are first, second and third number respectively, then according to question: x y + z = 26 and y = 3x

(2) (C) $S=26 x^2-208 x+676$
Explanation: Since, x + y + z = 26 and y = 3x
Then, 4x + z = 26
$z=26-4 x$
Sum of the squares of three numbers,
$\begin{aligned} 5 & =x^2+y^2+x^2 \\ & =x^2+\left(3 x^2\right)+(26-4 x)^2 \\ & =x^2+9 x^2+676+16 x^2-208 x\end{aligned}$
$\Rightarrow \quad S=26 x^2-208 x+676$

(3) (A) 52
Explanation: Since $S=26 x^2-208 x+676$
On differentiating both sides w.r.t. x, we get
$\frac{d S}{d x}=52 x-208$
Now, $\frac{d^2 S}{d x^2}=\frac{d}{d x}\left(\frac{d S}{d x}\right)$
$=\frac{d}{d x}(52 x-208)=52$

(4) (D) 4
Explanation: For maxima or minima,
put $\frac{d S}{d x}=0$
$\Rightarrow$ 52x - 208 = 0
$\Rightarrow$ 52x = 208
$\therefore$ x = 4
Now, at x = 4 $\frac{d^2 S}{d x^2}=52>0$
Hence, at x = 4 S becomes minimum.

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Applied Maths Case study (4 Marks)

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